Actually the problem is wrong. Here is the given solution: By the Phytagorean theorem: $MA=1$. Now construct in the exteror of $\triangle ABC$ the triangle $\triangle BNE \equiv \triangle BMA$. Now, triangle $\triangle BNM$ is equilateral and hence $MN= \sqrt{2}$. From the reciprocal of Phytagorean theorem it follows now that $\angle MNE\equiv 90$, so after summing we get $\angle BNE \equiv150$. Now, by the Cosine theorem it follows that $AE^2=3+\sqrt{6}$ and this also represents the area to find. But, if we search only for a square where exists a point $M$ inside it such that $MB^2=2$, $MC^2=3$, $MD^2=5$ using carteaian coordinates, with $A(0,0)$, $B(a,0)$, $C(a,a)$, $D(0,a)$ and $M(x,y)$, solving the system formed by the equations of distances we find that there exists only two squares, with $a=7+/- \sqrt{15}$, and that squares don’t verify the condition $ME^2=3$. Conclusion: if we put initially the condition $ABCD$ square we find no solution to the problem. But, analysing their solution we see that the only condition about $ABCD$ used was that it is a rectangle. Therefore, we can state the problem in the following way: Correct problem: Let $ABCD$ be a rectangle with $AD\geq AB$ and an equilateral triangle $\triangle ABE$ inside it is constructed. It is known that there exists a point $M$ inside $\triangle ABE$ such that $MB^2=2$, $MC^2=6$, $MD^2=5$, $ME^2=3$. Find the lengths of $AB$ and $AD$. We can see that $AB$ will be found in the same way, so in this problem also $AB^2= 3+\sqrt{6}$. Now, $AD$ will be different. And hence the area of the rectangle will be different.